Question

A current of 9.65 ampere is passed through the aqueous solution NaCl using suitable electrodes for 1000 s. The amount of NaOH formed during electrolysis is

• A. 2.0 g
• B. 4.0 g
• C. 6.0 g
• D. 8.0 g

Answer 1

Answer: (B)

4.0 g

Answer 2

2H2O + 2e– $\longrightarrow$ H2 + 2OH–

No. of Faraday passed = $\frac { 9.65 \times 1000 } { 96500 }$ = 0.1 F

$\mathrm { n } _ { \mathrm { OH } ^ { - } }$formed = 0.1 mol

nNaOH = 0.1 mol 4 gm.