To deposit one equivalent weight of silver at cathode the charge required will be

Question

To deposit one equivalent weight of silver at cathode the charge required will be (A) 9.65 × 104 C (B)  9.65× 103 C  (C)  9.65× 105 C  (D)  9.65× 107 C

Tardigrade Admin · Accepted Answer

One faraday is that quantity of electricity which can be deposited or liberated weight equal to equivalent weight.

1

faraday

= 96500 C

Let the required charge to deposit one equivalent weight of silver be

Q

.
Equivalent weight

= \frac{Q}{F}

1 = \frac{Q}{96500} (1 F = 96500 C)

Q = 96500 C = 9.65 \times 1 0^{4} C

Q. To deposit one equivalent weight of silver at cathode the charge required will be

$9.65 \times 1 0^{4} C$

$9.65 \times 1 0^{3} C$

$9.65 \times 1 0^{5} C$

$9.65 \times 1 0^{7} C$

Q. To deposit one equivalent weight of silver at cathode the charge required will be

9.65×104C

9.65×103C

9.65×105C

9.65×107C

One faraday is that quantity of electricity which can be deposited or liberated weight equal to equivalent weight. 1 faraday =96500C Let the required charge to deposit one equivalent weight of silver be Q. Equivalent weight =FQ​ 1=96500Q​(1F=96500C) Q=96500C=9.65×104C

$9.65 \times 1 0^{4} C$

$9.65 \times 1 0^{3} C$

$9.65 \times 1 0^{5} C$

$9.65 \times 1 0^{7} C$