A solution of $N i (N O_{3})_{2}$ is electrolysed between platinum electrodes using a current of 5 amperes for 20 min. What mass of Ni is deposited at the cathode ? (Atomic mass of Ni = 58.7)

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If 0.50 L of a 0.60 M SnSO4 solution is electrolysed for a period of 30.0 min using a current of 4.60 A. If in electrodes are used, what is the final concentration of Sn2+ remaining in the solution? [at.wt. of Sn=119]

The correct option is A 0.257 MAns : AGiven : 0.6 mol in 1 L then in 0.5 L = 0.3 mol of Sn2+ Mass deposited W = zit =1192×96500×4.6×1800 = 5.1 g Rest mass in solution = 30.6 gm Now molority of solution = 30.6119×1 = o.257 M

If $0.50 L$ of a $0.60 M S n S O_{4}$ solution is electrolysed for a period of $30.0$ min using a current of $4.60 A$ . If in electrodes are used, what is the final concentration of $S n^{2 +}$ remaining in the solution? $[a t . w t . o f S n = 119]$

The correct option is A $0.257 M$
Ans : A
Given : 0.6 mol in 1 L
then in 0.5 L = 0.3 mol of $S n^{2 +}$
Mass deposited W = zit
= $\frac{119}{2 \times 96500} \times 4.6 \times 1800$
= 5.1 g
Rest mass in solution = 30.6 gm
Now molority of solution = $\frac{30.6}{119 \times 1}$
= o.257 M