Question

The time taken by the galvanic cell which operates almost ideally under reversible conditions at a current of ${10}^{-16}A$ to deliver 1 mole of electron is :

• A. $19.3\times {10}^{-17}s$
• B. $4.82\times {10}^{20}s$
• C. $9.65\times {10}^{20}s$
• D. $3.4\times {10}^{11}s$

Answer 1

The correct option is C $9.65\times {10}^{20}s$

Given :
$\text{Current}={10}^{-16}A$
$\text{Charge}=96500C$
( Charge on 1 mole of electrons)

Using the formula,
$\text{Charge (Q)}=\text{Current (A)}\times \text{Time (T)}$

Therefore,
$T=\frac{96500}{{10}^{-16}}$

$T=9.65\times {10}^{20}s$