Question

When the same amount of electricity is passed through solution of silver nitrate and copper sulfate, $0.4$ g copper is deposited.

The amount of silver deposited is:

• A. $1.36$g
• B. $2.7$g
• C. $5.1$g
• D. $5.4$g

Answer 1

The correct option is A $1.36$g

$\text{moles of Cu deposited}=\frac{\text{mass}}{\text{molar mass}}$
$\text{moles of Cu deposited}=\frac{\text{0.4 g}}{\text{63.5 g/mol}}$
$\text{moles of Cu deposited}=\text{0.006299 moles}$
The half reactions for the formation of Ag and Cu are
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$
mole ratio $=\frac{\text{1 mol Ag}}{\text{1 mol}{e}^{-}}$
$C{u}^{++}\left(aq\right)+2e^{-}\to Cu\left(s\right)$
mole ratio $=\frac{\text{1 mol Cu}}{\text{2 mol}{e}^{-}}$
$\frac{\text{moles of Cu deposited}}{\text{moles of Ag deposited}}=\frac{\text{mole ratio of Cu half reaction}}{\text{mole ratio of Ag half reaction}}$
$\frac{\text{0.006299 moles}}{\text{moles of Ag deposited}}=\frac{\frac{\text{1 mol Cu}}{\text{2 mol}{e}^{-}}}{\frac{\text{1 mol Ag}}{\text{1 mol}{e}^{-}}}$
$\frac{\text{0.006299 moles}}{\text{moles of Ag deposited}}=\frac{1}{2}$

$\text{moles of Ag deposited}=2\times \text{0.006299 moles}$
$\text{moles of Ag deposited}=\text{0.0126 moles}$
The molar mass of $Ag$ is $108g/mol$.
$\text{Mass of Ag deposited}=\text{0.0126 moles}\times 108g/mol=\text{1.36 g}$