Question

Given the equilibrium constant, $K_c$ of the reaction :

$Cu\left(s\right)+2A{g}^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2Ag\left(s\right)$ is $10\times {10}^{15}$, calculate the $E_{cell}^o$ of this reaction at $298K$

$[2.303\frac{RT}{F}at298K=0.059V]$

• A. $0.04736V$
• B. $0.4736V$
• C. $0.4736mV$
• D. $0.04736mV$

Answer 1

The correct option is B $0.4736V$

$E_{cell}=E_{cell}^o-\frac{0.059}{n}logQ$

At equilibrium,

$E_{cell}^o=\frac{0.059}{n}logK$

$E_{cell}^o=\frac{0.059}{2}log(10\times {10}^{15})$

$E_{cell}^o=0.059\times 8$

$E_{cell}^o=0.472V$