What will be the reduction potential for the following half-cell reaction at 298 K ? (Given: [Ag+]=0.1 M and E°cell=+0.80 V)

Question

What will be the reduction potential for the following half-cell reaction at 298 K ? (Given: [Ag+]=0.1 M and E°cell=+0.80 V) (A) 0.741 V (B) 0.80 V (C) -0.80 V (D) -0.741 V

Tardigrade Admin · Accepted Answer

Ag++e- → Ag Ecell=E°cell-(0.0591/1) log (1/[Ag+]) Ecell=0.80-(0.0591/1) log (1/0.1) =0.80-0.0591 =0.741 V

Q. What will be the reduction potential for the following half-cell reaction at $298 K$ ?
(Given : $[A g^{+}] = 0.1 M$ and $E_{ce ll}^{\circ} = + 0.80 V$ )

$0.741 V$

$0.80 V$

$- 0.80 V$

$- 0.741 V$

$A g^{+} + e^{-} \to A g$
$E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.0591}{1} l o g \frac{1}{[ A g ^{+} ]}$
$E_{ce ll} = 0.80 - \frac{0.0591}{1} l o g \frac{1}{0.1}$
$= 0.80 - 0.0591$
$= 0.741 V$

Q. What will be the reduction potential for the following half-cell reaction at 298K ? (Given : [Ag+]=0.1M and Ecell∘​=+0.80V)

0.741V

0.80V

−0.80V

−0.741V