Question

The difference between standard electrode potential (reduction) of metal-metal insoluble salts of AgCl and AgBr is x. Calculate the ratio of the solubility product of AgCl and AgBr.

• A. $antilog\left[\frac{x}{0.059}\right]$
• B. $antilog\left[\frac{x}{2\times 0.059}\right]$
• C. $antilog\left[\frac{4x}{0.059}\right]$
• D. None of these

Answer 1

The correct option is A $antilog\left[\frac{x}{0.059}\right]$

$E_{cell}=0=E_{cell}^0-\frac{0.059}{1}\log K_{sp}\left(AgCl\right)$
or $E_{cell}^0=\frac{0.059}{1}\log K_{sp}\left(AgCl\right)$ ........(A)
Similarly for $AgBr{E}_{cell}^{{}^{\prime }0}=\frac{0.059}{1}\log K_{sp}\left(AgBr\right)$ ........(B)
On applying (A) - (B), $x=\frac{0.059}{1}\log \frac{K_{sp}\left(AgCl\right)}{K_{sp}\left(AgBr\right)}$