1/2Zn^2+ (aq) + e^- 一> 1/2Zn (s). In the reaction above, calculate the quantity of electricity required to discharge zinc?

1/2Zn^2+ (aq) + e^- 一> 1/2Zn (s). In the reaction above, calculate the quantity of electricity required to discharge zinc? [F=96500CMol-1] 

A. 0.965*10^4 C 

 B. 4.820*10^4 C 

 C. 9.650*10^4 C 

 D. 48.200*10^4 C

Given half-reaction:

Zn^{2+}(aq) + e^- 一> Zn (s)

This equation shows that 1 mole of electrons is required.

Using Faraday’s law:
Q = nF

Where
(n = 1) mole of electrons
(F = 96,500 \C mol^{-1}

Q = 1 * 96,500 = 9.650 * 10^4 C

✅ Correct answer: C. (9.650 * 10^4 C

PQN: 15