Question

Silver can be plated out of a solution containing Ag+ according to the half-reaction
Ag+(aq)+e−→Ag(s)


How much time (in minutes) does it take to plate 19 g of silver using a current of 3.4 A ?

Answer 1

Answer:

Approximately $83$ minutes.

Explanation:

Look up the relative atomic mass of $\rm Ag$: $M({\rm Ag}) = 107.868\; \rm g \cdot mol^{-1}$.$\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 1.6987 \times 10^{4}\; \rm C \end{aligned}$/.

Avogadro's number: $N_A \approx 6.02 \times 10^{23}\; \rm mol^{-1}$.

Elementary charge: $e \approx 1.602 \times 10^{-19}\; \rm C$.

Calculate the quantity of $\rm Ag$ atoms to reduce:

$\begin{aligned}& n({\rm Ag}) \\ &= \frac{m({\rm Ag})}{M({\rm Ag})} \\ &= \frac{19\; \rm g}{107.868\; \rm g \cdot mol^{-1}} \\ & \approx 0.176\; \rm mol\end{aligned}$.

By the equation, it takes one electron to reduce every $\rm Ag$ atom. Thus, the number of electrons required to reduce $0.176\; \rm mol$ of $\rm Ag\!$ atoms would be:

$n(e) = n({\rm Ag}) \approx 0.176\; \rm mol$.

$\begin{aligned}N(e) &= n(e) \cdot N_{A}. \\ &\approx 0.176\; \rm mol \times 6.02 \times 10^{23}\; \rm mol^{-1} \\ & \approx 1.06 \times 10^{23}\end{aligned}$.

Calculate the amount of charge (in coulombs) in that many electrons:

$\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 16987.1 \; \rm C \end{aligned}$.

A current of $1\; \rm A$ carries a charge of $1\; \rm C$ every second. Thus, the amount of time required for this current to carry that much electron would be:"

$\begin{aligned}t &= \frac{Q}{I} \\ &\approx \frac{16987.1\; \rm C}{3.4\; \rm A} \\ &\approx 83.3\; \rm s \\ &\approx 5.00\times 10^{3}\; \rm s \\ &\approx 83\; \text{minutes} \end{aligned}$.