275780
A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathbf{p H}$ $=9$ and passing hydrogen gas around the platinum wire at $1.2 \mathrm{~atm}$ pressure. The oxidation potential of such an eletrode equals V.
1 +0.59
2 -0.531
3 -0.59
4 +0.531
Explanation:
$\mathrm{pH}$ of $\mathrm{HNO}_{3}$ solution $=9$ Then, $\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-9} \mathrm{M}$ According to Nernst equation for hydrogen electrode- $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\left(\mathrm{H}^{+}\right)}{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)}\left\{\begin{array}{l} \mathrm{n}=1 \\ \text { For hydrogen } \\ \text { electrode i.e. } \mathrm{E}^{\circ}=0 \end{array}\right\} \\ & \mathrm{E}_{\text {cell }}=0-\frac{0.059}{1} \log \frac{10^{-9}}{1} \\ & \mathrm{E}_{\text {cell }}=-0.059 \times(-9) \\ & \mathrm{E}_{\text {cell }}=0.531 \end{aligned}$
AP-EAMCET 25-08-2021
ELECTROCHEMISTRY
275781
For the cell reaction $\mathrm{Cu} \mid \mathrm{Cu}^{2+}(0.1 \mathrm{M})$ $\ \vert \mathrm{Cu}^{2+}(1.0 \mathrm{M}) \mid \mathrm{Cu}$; the emf of the cell at $25^{\circ} \mathrm{C}$ is $\left[\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\right]$
1 $0.059 \mathrm{~V}$
2 $0.311 \mathrm{~V}$
3 $0.369 \mathrm{~V}$
4 $0.029 \mathrm{~V}$
Explanation:
Given, $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}$ We can write the Nernst equation for the given cell- $\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log \frac{[0.1]}{[1.0]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log 10^{-1} \\ & \mathrm{E}=0+\frac{0.059}{2} \\ & \mathrm{E}=0.029 \mathrm{~V} \end{aligned}$
TS-EAMCET (Engg.)
ELECTROCHEMISTRY
275775
Which is symbolic representation for following cell reaction, $\mathrm{Mg}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}$.
275783
Given half-cell potentials $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{0}=0.4 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{0}=0.91 \mathrm{~V}$. Find the standard reduction potential of $\mathrm{Cr}^{3+} / \mathrm{Cr}$
275780
A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathbf{p H}$ $=9$ and passing hydrogen gas around the platinum wire at $1.2 \mathrm{~atm}$ pressure. The oxidation potential of such an eletrode equals V.
1 +0.59
2 -0.531
3 -0.59
4 +0.531
Explanation:
$\mathrm{pH}$ of $\mathrm{HNO}_{3}$ solution $=9$ Then, $\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-9} \mathrm{M}$ According to Nernst equation for hydrogen electrode- $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\left(\mathrm{H}^{+}\right)}{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)}\left\{\begin{array}{l} \mathrm{n}=1 \\ \text { For hydrogen } \\ \text { electrode i.e. } \mathrm{E}^{\circ}=0 \end{array}\right\} \\ & \mathrm{E}_{\text {cell }}=0-\frac{0.059}{1} \log \frac{10^{-9}}{1} \\ & \mathrm{E}_{\text {cell }}=-0.059 \times(-9) \\ & \mathrm{E}_{\text {cell }}=0.531 \end{aligned}$
AP-EAMCET 25-08-2021
ELECTROCHEMISTRY
275781
For the cell reaction $\mathrm{Cu} \mid \mathrm{Cu}^{2+}(0.1 \mathrm{M})$ $\ \vert \mathrm{Cu}^{2+}(1.0 \mathrm{M}) \mid \mathrm{Cu}$; the emf of the cell at $25^{\circ} \mathrm{C}$ is $\left[\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\right]$
1 $0.059 \mathrm{~V}$
2 $0.311 \mathrm{~V}$
3 $0.369 \mathrm{~V}$
4 $0.029 \mathrm{~V}$
Explanation:
Given, $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}$ We can write the Nernst equation for the given cell- $\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log \frac{[0.1]}{[1.0]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log 10^{-1} \\ & \mathrm{E}=0+\frac{0.059}{2} \\ & \mathrm{E}=0.029 \mathrm{~V} \end{aligned}$
TS-EAMCET (Engg.)
ELECTROCHEMISTRY
275775
Which is symbolic representation for following cell reaction, $\mathrm{Mg}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}$.
275783
Given half-cell potentials $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{0}=0.4 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{0}=0.91 \mathrm{~V}$. Find the standard reduction potential of $\mathrm{Cr}^{3+} / \mathrm{Cr}$
275780
A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathbf{p H}$ $=9$ and passing hydrogen gas around the platinum wire at $1.2 \mathrm{~atm}$ pressure. The oxidation potential of such an eletrode equals V.
1 +0.59
2 -0.531
3 -0.59
4 +0.531
Explanation:
$\mathrm{pH}$ of $\mathrm{HNO}_{3}$ solution $=9$ Then, $\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-9} \mathrm{M}$ According to Nernst equation for hydrogen electrode- $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\left(\mathrm{H}^{+}\right)}{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)}\left\{\begin{array}{l} \mathrm{n}=1 \\ \text { For hydrogen } \\ \text { electrode i.e. } \mathrm{E}^{\circ}=0 \end{array}\right\} \\ & \mathrm{E}_{\text {cell }}=0-\frac{0.059}{1} \log \frac{10^{-9}}{1} \\ & \mathrm{E}_{\text {cell }}=-0.059 \times(-9) \\ & \mathrm{E}_{\text {cell }}=0.531 \end{aligned}$
AP-EAMCET 25-08-2021
ELECTROCHEMISTRY
275781
For the cell reaction $\mathrm{Cu} \mid \mathrm{Cu}^{2+}(0.1 \mathrm{M})$ $\ \vert \mathrm{Cu}^{2+}(1.0 \mathrm{M}) \mid \mathrm{Cu}$; the emf of the cell at $25^{\circ} \mathrm{C}$ is $\left[\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\right]$
1 $0.059 \mathrm{~V}$
2 $0.311 \mathrm{~V}$
3 $0.369 \mathrm{~V}$
4 $0.029 \mathrm{~V}$
Explanation:
Given, $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}$ We can write the Nernst equation for the given cell- $\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log \frac{[0.1]}{[1.0]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log 10^{-1} \\ & \mathrm{E}=0+\frac{0.059}{2} \\ & \mathrm{E}=0.029 \mathrm{~V} \end{aligned}$
TS-EAMCET (Engg.)
ELECTROCHEMISTRY
275775
Which is symbolic representation for following cell reaction, $\mathrm{Mg}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}$.
275783
Given half-cell potentials $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{0}=0.4 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{0}=0.91 \mathrm{~V}$. Find the standard reduction potential of $\mathrm{Cr}^{3+} / \mathrm{Cr}$
275780
A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathbf{p H}$ $=9$ and passing hydrogen gas around the platinum wire at $1.2 \mathrm{~atm}$ pressure. The oxidation potential of such an eletrode equals V.
1 +0.59
2 -0.531
3 -0.59
4 +0.531
Explanation:
$\mathrm{pH}$ of $\mathrm{HNO}_{3}$ solution $=9$ Then, $\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-9} \mathrm{M}$ According to Nernst equation for hydrogen electrode- $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\left(\mathrm{H}^{+}\right)}{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)}\left\{\begin{array}{l} \mathrm{n}=1 \\ \text { For hydrogen } \\ \text { electrode i.e. } \mathrm{E}^{\circ}=0 \end{array}\right\} \\ & \mathrm{E}_{\text {cell }}=0-\frac{0.059}{1} \log \frac{10^{-9}}{1} \\ & \mathrm{E}_{\text {cell }}=-0.059 \times(-9) \\ & \mathrm{E}_{\text {cell }}=0.531 \end{aligned}$
AP-EAMCET 25-08-2021
ELECTROCHEMISTRY
275781
For the cell reaction $\mathrm{Cu} \mid \mathrm{Cu}^{2+}(0.1 \mathrm{M})$ $\ \vert \mathrm{Cu}^{2+}(1.0 \mathrm{M}) \mid \mathrm{Cu}$; the emf of the cell at $25^{\circ} \mathrm{C}$ is $\left[\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\right]$
1 $0.059 \mathrm{~V}$
2 $0.311 \mathrm{~V}$
3 $0.369 \mathrm{~V}$
4 $0.029 \mathrm{~V}$
Explanation:
Given, $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}$ We can write the Nernst equation for the given cell- $\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log \frac{[0.1]}{[1.0]} \\ & \mathrm{E}=0-\frac{0.059}{2} \log 10^{-1} \\ & \mathrm{E}=0+\frac{0.059}{2} \\ & \mathrm{E}=0.029 \mathrm{~V} \end{aligned}$
TS-EAMCET (Engg.)
ELECTROCHEMISTRY
275775
Which is symbolic representation for following cell reaction, $\mathrm{Mg}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}$.
275783
Given half-cell potentials $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{0}=0.4 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{0}=0.91 \mathrm{~V}$. Find the standard reduction potential of $\mathrm{Cr}^{3+} / \mathrm{Cr}$
