Question

For the cell reaction,

$Mg+2{Ag}^{+}\left(0.0001M\right)\to {Mg}^{2+}\left(0.01M\right)+2Ag$

$E_{cell}^o=3.177V$ then what will be value of $E_{cell}$ for the above reaction?

• A. $2.5$
• B. $3.0$
• C. $3.2$
• D. $3.8$

Answer 1

The correct option is B $3.0$

$Mg+2{Ag}^{+}\left(0.0001M\right)\to {Mg}^{2+}\left(0.01M\right)+2Ag$

$E_{cell}=E^0-\frac{0.059}{n}log\left[\frac{\left({Mg}^{2+}\right)}{{\left({Ag}^{+}\right)}^2}\right]$

$=3.177-\frac{0.059}{2}log\left[\frac{0.01}{{\left(0.0001\right)}^2}\right]$

$=3.177-\frac{0.059}{2}\times 6$

$=3.177-0.177$

$E_{cell}=3.0V$

Hence, option B is correct.