Question

The standard emf of a cell, involving one electron change is found to be $0.591V$ at $25$. The equilibrium constant of the reaction is $(F=96500C{mol}^{-1},R=8.314K^{-1}{mol}^{-1})$

• A. $1.0\times {10}^1$
• B. $1.0\times {10}^5$
• C. $1.0\times {10}^{10}$
• D. $1.0\times {10}^{30}$

Answer 1

The correct option is C $1.0\times {10}^{10}$

For a cell Reaction in equilibrium at 298 K ,

$E_{cell}^o=\frac{0.0591}{n}log{K}_c$

$(K_c=$ $equilibrium$ $constant)$

Give, $E_{cell}^o=0.0591V$

Now, $log{K}_c=\frac{E_{cell}^o\ast n}{0.0591}$

$=\frac{0.0591\ast n}{0.0591}$

$log{K}_c=10$

$K_c=antilog10$

$K_c=1\ast {10}^{10}$

therefore the Answer is option C.