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CHEMISTRY(KCET)

285409 For ${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O, E^{\circ} =$ $1.33 V At [{Cr}_{2} O_{7}^{2 -}] = 4.5$ millimole, $[{Cr}^{3 +}] = 15$ millimole, E is 1.067 V . The pH of the solution is nearly equal to

1 2

2 3

3 5

4 4

Explanation:

(a) For the given reaction.
$\begin{aligned} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O \\ E = E^{\circ} - \frac{2.303 RT}{nF} \log \frac{[Products]}{[Reactants]} \\ 1.067 = 1.33 - \frac{0.0591}{6} \log \frac{{[{Cr}^{3 +}]}^{2} {[H_{2} O]}^{7}}{[{Cr}_{2} O_{7}^{2 -}] {[H^{+}]}^{14}} \\ 1.067 = 1.33 - 9.85 \times 10^{- 3} \\ \log \frac{{[15 \times 10^{- 3}]}^{2} [1]^{7}}{[4.5 \times 10^{- 3}] {[H^{+}]}^{14}} \\ \frac{- 0.263}{26} = 2 \log [15 \times 10^{- 3}] \\ - 9.85 \times 10^{- 3} \\ 26.7 = 2 \times - 1.82 + 2.34 + 14 pH \\ 26.7 = - 3.64 + 2.34 + 14 pH \\ 14 pH = 26.7 + 3.64 - 2.34 \\ pH = \frac{28}{14} = 2 \end{aligned}$

Karnataka CET 2014

CHEMISTRY(KCET)

285410 Impure copper containing $Fe, Au, Ag$ as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g . Percentage of iron in impure copper is
(Given : molar mass $Fe = 55.5 g {mol}^{- 1}$ , molar mass $Cu = 63.54 g {mol}^{- 1}$ )

1 0.95

2 0.85

3 0.97

4 0.90

Explanation:

(d) ${Cu}^{2 +} + 2 e^{-} \to Cu$
Equivalent wt. $= \frac{63.54}{2} = 31.77$
${Fe}^{2 +} + 2 e^{-} \to Fe$
Equivalent wt. $= \frac{55.5}{2} = 27.75$
Mass increased at cathode is due to deposition of Cu.
Hence, no. of gram equivalents of Cu deposited
$= \frac{22.011}{31.77} = 0.6928$
Now, using Faraday's first law,
$Q = it = 140 A \times 482.5 s = 67550 C$
96,500 C gives 31.77 g Cu
67550 C will give $= \frac{31.77}{96, 500} \times \frac{67550}{1} = 22.239 g$ of pure
Cu
But, according to the question, the mass of cathode only increases $= 22.011 g$
Hence, $22.239 - 22.011 = 0.228 g$
Now, $31.77 g Cu \equiv 0.228 g$ deposited
$27.75 g Fe \equiv \frac{0.228}{31.77} \times 27.75$
$= 0.199 g$ Fe deposited
$%$ of $Fe = \frac{Mass of Fe}{Mass of impurities} \times 100$
$= \frac{0.199}{22.26} \times 100 = 0.894 \approx 0.90 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285411 For hydrogen - oxygen fuel cell at one atm and 298 K
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ G^{\circ} = - 240 kJ$
$E^{\circ}$ for the cell is approximately,

(Given : $F = 96, 500 C$ )

1 2.48 V

2 1.24 V

3 2.5 V

4 1.26 V

Explanation:

(b) $Δ G^{\circ} = - nFE^{\circ}$
$- 240000 = - 2 \times 96, 500 \times E^{\circ}$
$\Rightarrow E^{\circ} = 1.24 V$

Karnataka CET 2014

CHEMISTRY(KCET)

285409 For ${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O, E^{\circ} =$ $1.33 V At [{Cr}_{2} O_{7}^{2 -}] = 4.5$ millimole, $[{Cr}^{3 +}] = 15$ millimole, E is 1.067 V . The pH of the solution is nearly equal to

1 2

2 3

3 5

4 4

Explanation:

(a) For the given reaction.
$\begin{aligned} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O \\ E = E^{\circ} - \frac{2.303 RT}{nF} \log \frac{[Products]}{[Reactants]} \\ 1.067 = 1.33 - \frac{0.0591}{6} \log \frac{{[{Cr}^{3 +}]}^{2} {[H_{2} O]}^{7}}{[{Cr}_{2} O_{7}^{2 -}] {[H^{+}]}^{14}} \\ 1.067 = 1.33 - 9.85 \times 10^{- 3} \\ \log \frac{{[15 \times 10^{- 3}]}^{2} [1]^{7}}{[4.5 \times 10^{- 3}] {[H^{+}]}^{14}} \\ \frac{- 0.263}{26} = 2 \log [15 \times 10^{- 3}] \\ - 9.85 \times 10^{- 3} \\ 26.7 = 2 \times - 1.82 + 2.34 + 14 pH \\ 26.7 = - 3.64 + 2.34 + 14 pH \\ 14 pH = 26.7 + 3.64 - 2.34 \\ pH = \frac{28}{14} = 2 \end{aligned}$

Karnataka CET 2014

CHEMISTRY(KCET)

285410 Impure copper containing $Fe, Au, Ag$ as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g . Percentage of iron in impure copper is
(Given : molar mass $Fe = 55.5 g {mol}^{- 1}$ , molar mass $Cu = 63.54 g {mol}^{- 1}$ )

1 0.95

2 0.85

3 0.97

4 0.90

Explanation:

(d) ${Cu}^{2 +} + 2 e^{-} \to Cu$
Equivalent wt. $= \frac{63.54}{2} = 31.77$
${Fe}^{2 +} + 2 e^{-} \to Fe$
Equivalent wt. $= \frac{55.5}{2} = 27.75$
Mass increased at cathode is due to deposition of Cu.
Hence, no. of gram equivalents of Cu deposited
$= \frac{22.011}{31.77} = 0.6928$
Now, using Faraday's first law,
$Q = it = 140 A \times 482.5 s = 67550 C$
96,500 C gives 31.77 g Cu
67550 C will give $= \frac{31.77}{96, 500} \times \frac{67550}{1} = 22.239 g$ of pure
Cu
But, according to the question, the mass of cathode only increases $= 22.011 g$
Hence, $22.239 - 22.011 = 0.228 g$
Now, $31.77 g Cu \equiv 0.228 g$ deposited
$27.75 g Fe \equiv \frac{0.228}{31.77} \times 27.75$
$= 0.199 g$ Fe deposited
$%$ of $Fe = \frac{Mass of Fe}{Mass of impurities} \times 100$
$= \frac{0.199}{22.26} \times 100 = 0.894 \approx 0.90 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285411 For hydrogen - oxygen fuel cell at one atm and 298 K
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ G^{\circ} = - 240 kJ$
$E^{\circ}$ for the cell is approximately,

(Given : $F = 96, 500 C$ )

1 2.48 V

2 1.24 V

3 2.5 V

4 1.26 V

Explanation:

(b) $Δ G^{\circ} = - nFE^{\circ}$
$- 240000 = - 2 \times 96, 500 \times E^{\circ}$
$\Rightarrow E^{\circ} = 1.24 V$

Karnataka CET 2014

CHEMISTRY(KCET)

285409 For ${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O, E^{\circ} =$ $1.33 V At [{Cr}_{2} O_{7}^{2 -}] = 4.5$ millimole, $[{Cr}^{3 +}] = 15$ millimole, E is 1.067 V . The pH of the solution is nearly equal to

1 2

2 3

3 5

4 4

Explanation:

(a) For the given reaction.
$\begin{aligned} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O \\ E = E^{\circ} - \frac{2.303 RT}{nF} \log \frac{[Products]}{[Reactants]} \\ 1.067 = 1.33 - \frac{0.0591}{6} \log \frac{{[{Cr}^{3 +}]}^{2} {[H_{2} O]}^{7}}{[{Cr}_{2} O_{7}^{2 -}] {[H^{+}]}^{14}} \\ 1.067 = 1.33 - 9.85 \times 10^{- 3} \\ \log \frac{{[15 \times 10^{- 3}]}^{2} [1]^{7}}{[4.5 \times 10^{- 3}] {[H^{+}]}^{14}} \\ \frac{- 0.263}{26} = 2 \log [15 \times 10^{- 3}] \\ - 9.85 \times 10^{- 3} \\ 26.7 = 2 \times - 1.82 + 2.34 + 14 pH \\ 26.7 = - 3.64 + 2.34 + 14 pH \\ 14 pH = 26.7 + 3.64 - 2.34 \\ pH = \frac{28}{14} = 2 \end{aligned}$

Karnataka CET 2014

CHEMISTRY(KCET)

285410 Impure copper containing $Fe, Au, Ag$ as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g . Percentage of iron in impure copper is
(Given : molar mass $Fe = 55.5 g {mol}^{- 1}$ , molar mass $Cu = 63.54 g {mol}^{- 1}$ )

1 0.95

2 0.85

3 0.97

4 0.90

Explanation:

(d) ${Cu}^{2 +} + 2 e^{-} \to Cu$
Equivalent wt. $= \frac{63.54}{2} = 31.77$
${Fe}^{2 +} + 2 e^{-} \to Fe$
Equivalent wt. $= \frac{55.5}{2} = 27.75$
Mass increased at cathode is due to deposition of Cu.
Hence, no. of gram equivalents of Cu deposited
$= \frac{22.011}{31.77} = 0.6928$
Now, using Faraday's first law,
$Q = it = 140 A \times 482.5 s = 67550 C$
96,500 C gives 31.77 g Cu
67550 C will give $= \frac{31.77}{96, 500} \times \frac{67550}{1} = 22.239 g$ of pure
Cu
But, according to the question, the mass of cathode only increases $= 22.011 g$
Hence, $22.239 - 22.011 = 0.228 g$
Now, $31.77 g Cu \equiv 0.228 g$ deposited
$27.75 g Fe \equiv \frac{0.228}{31.77} \times 27.75$
$= 0.199 g$ Fe deposited
$%$ of $Fe = \frac{Mass of Fe}{Mass of impurities} \times 100$
$= \frac{0.199}{22.26} \times 100 = 0.894 \approx 0.90 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285411 For hydrogen - oxygen fuel cell at one atm and 298 K
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ G^{\circ} = - 240 kJ$
$E^{\circ}$ for the cell is approximately,

(Given : $F = 96, 500 C$ )

1 2.48 V

2 1.24 V

3 2.5 V

4 1.26 V

Explanation:

(b) $Δ G^{\circ} = - nFE^{\circ}$
$- 240000 = - 2 \times 96, 500 \times E^{\circ}$
$\Rightarrow E^{\circ} = 1.24 V$

Karnataka CET 2014