Question

For the following electrochemical cell at 298K, $Pt\left(s\right)|H_2(g,1bar)|H^{+}(aq,1M)||M^{4+}\left(aq\right),M^{2+}\left(aq\right)|Pt\left(s\right)$

$E_{cell}=0.092$V when $\frac{\left[M^{2+}\right(aq\left)\right]}{\left[M^{4+}\right(aq\left)\right]}={10}^x$.

Given $:E_{M^{4+}/M^{2+}}^0=0.151V;2.303\frac{RT}{F}=0.059V$

The value of x is :

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

Given:

$E_{cell}=0.092$V, $\frac{\left[M^{2+}\right(aq\left)\right]}{\left[M^{4+}\right(aq\left)\right]}={10}^x$.

$p{H}_2=1bar$

$E_{\text{cell}}=E_{cell}^o-\frac{0.059}{2}lo{g}_{10}\frac{\left[M^{2+}\right][H^{+}{]}^2}{\left[M^{4+}\right]p{H}_2}$

$0.092=0.151-\frac{0.059}{2}lo{g}_{10}{10}^x$

$\therefore x=2$

Hence, the correct option is D.