Question

Silver and a zinc voltmeter are connected in a series and a current is passed through them for a time t, liberating W g of zinc. The weight of silver deposited is nearly:
A) \[Wg\]
B) $1.7Wg$
C) $2.4Wg$
D) $3.3Wg$

Answer 1

Hint: We know that, when the two conductors are connected in series, the current passes through both of the conductors is the same. We can use the formulae of metal liberated due to the current flow to find the amount of zinc and silver liberated separately, and then using the amounts we can get our answers.

Formula used:
To find the amount of metal liberated, we will the following formula
 ${W_m} = \dfrac{{{w_m}}}{n}\left( {\dfrac{{I \cdot t}}{F}} \right)$
where ${w_m}$ is the atomic weight of the metal, \[n\] is the valency of the metal, $I$ is the current passing through metals, $t$ is the time.

Complete step by step solution:
Now, we have to find the amount of zinc and silver liberated separately.
Now, using the formula
${W_m} = \dfrac{{{w_m}}}{n}\left( {\dfrac{{I \cdot t}}{F}} \right)$

Let ${W_s}$ be the amount of silver liberated. The amount of silver liberated will be,
${W_s} = \dfrac{{{w_s}}}{n}\left( {\dfrac{{I \cdot t}}{F}} \right).........\left( 1 \right)$
We know that ${w_s}$ is the atomic weight of silver which is approximately $108$ and \[n\] is the valency of silver which is $1$.

Now, by substituting the values in the equation (1)
We get,
$
   \Rightarrow {W_s} = \dfrac{{{w_s}}}{n}\left( {\dfrac{{I \cdot t}}{F}} \right) \\
   \Rightarrow {W_s} = \dfrac{{108}}{1}\left( {\dfrac{{I \cdot t}}{F}} \right) \\
   \Rightarrow {W_s} = 108\left( {\dfrac{{I \cdot t}}{F}} \right).......(2) \\
 $
Let ${W_z}$ be the amount of zinc liberated. The amount of Zinc liberated will be,
${W_z} = \dfrac{{{w_z}}}{n}\left( {\dfrac{{I \cdot t}}{F}} \right).........\left( 3 \right)$
We know that ${w_z}$ is the atomic weight of silver which is approximately $65.38$ and \[n\] is the valency of zinc which is $2$.

Now, by substituting the values in the equation (3)
We get,

$
   \Rightarrow {W_z} = \dfrac{{{w_z}}}{n}\left( {\dfrac{{I \cdot t}}{F}} \right) \\
   \Rightarrow {W_z} = \dfrac{{65.38}}{2}\left( {\dfrac{{I \cdot t}}{F}} \right) \\
   \Rightarrow {W_z} = 32.69\left( {\dfrac{{I \cdot t}}{F}} \right).......(4) \\
 $
Now, dividing equation (2) by equation (4), we get,

$\dfrac{{{W_s}}}{{{W_z}}} = \dfrac{{108}}{{32.69}}$

Now, we have given in the question that ${W_z} = W$
So, putting this in the above equation, we get,

$
   \Rightarrow \dfrac{{{W_s}}}{W} = \dfrac{{108}}{{32.69}} \\
   \Rightarrow {W_s} = 3.3W \\
 $
Hence, the amount of silver liberated is $3.3Wg$.

Hence, the correct answer is option (D).

Note: We should always be careful in doing these types of questions. As these questions are always a bit confusing, hence, it is a little bit tougher to get a correct answer. Do the calculations very carefully, so that we can get an appropriate answer.