For the redox reaction Zn(s) +Cu2+ (0.1 M) - Zn2+ (1 M) +Cu(s) taking place in a cell, E°cell is 1.10 V . Ecell for the cell will be (2.303 (RT/F)=0.0591)

Question

For the redox reaction Zn(s) +Cu2+ (0.1 M) -> Zn2+ (1 M) +Cu(s) taking place in a cell, E°cell is 1.10 V . Ecell for the cell will be (2.303 (RT/F)=0.0591)  (A)  2.41 V  (B)  1.80 V  (C)  1.07 V  (D)  0.82 V

Tardigrade Admin · Accepted Answer

Ecell=E°cell-(0.0591/n) log Q underset text0.1 MCu2+ + Zn -> underset text1 MZn2+ + Cu Q=([Zn2+]/[Cu2+])=(1/0.1)=10 Ecell=1.10-(0.0591/2) log 10 = 1.10 â 0.0295 = 1.0705 V

Q. For the redox reaction
$Zn (s) + Cu X^{2 +} (0.1 M) Zn X^{2 +} (1 M) + Cu (s)$ taking place in a cell, $E_{ce ll}^{\circ}$ is $1.10 V$ . $E_{ce ll}$ for the cell will be $(2.303 \frac{RT}{F} = 0.0591)$

$2.41 V$

$1.80 V$

$1.07 V$

$0.82 V$

$E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.0591}{n} l o g Q$
$0.1 M Cu X^{2 +} + Zn$ $1 M Zn X^{2 +} + Cu$
$Q = \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]} = \frac{1}{0.1} = 10$
$E_{ce ll} = 1.10 - \frac{0.0591}{2} l o g 10$
$= 1.10 - 0.0295$
$= 1.0705 V$

Q. For the redox reaction Zn(s)+CuX2+ (0.1M)​ZnX2+ (1M)+Cu(s) taking place in a cell, Ecell∘​ is 1.10V . Ecell​ for the cell will be (2.303FRT​=0.0591)

2.41V

1.80V

1.07V

0.82V