Question

A galvanic cell is set up from a zinc bar weighing 50 g and 1.0 Liter, 1.0 M, ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ solution. How long would the cell run, assuming it delivers a steady current of 1 ampere.
a) 48 hours
b) 41hours
c) 21 hours
d) 1hour

Answer 1

Hint: Here, first we have to calculate the charge of the reaction. Then we have to use the formula, $t = \dfrac{Q}{I}$ to calculate the time. In the formula, Q represents charge, I represent current and t is time.

Complete step by step answer:
Let’s first write the oxidation half reaction of zinc. Zinc gives two electrons to form zinc ions.
${\rm{Zn}} \to {\rm{Z}}{{\rm{n}}^{2 + }} + 2{e^ - }$
          …… (1)
Now, we have to write the reduction reaction of copper ions to form copper.
${\rm{C}}{{\rm{u}}^{2 + }} + 2{{\rm{e}}^ - } \to {\rm{Cu}}$
           …… (2)
On adding (1) and (2), the equation obtained is,
${\rm{Zn}} + {\rm{C}}{{\rm{u}}^{2 + }} \to {\rm{Z}}{{\rm{n}}^{2 + }} + {\rm{Cu}}$
Now, we have to calculate the moles of ${\rm{C}}{{\rm{u}}^{{\rm{2 + }}}}$ ion and zinc ion.
Given that, the strength of ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ solution is 1.0 M and the mass of the copper sulphate solution is 1.0 Liter.
So, we can say that,
1 M ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ contains ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ =1 mole=1 mole ${\rm{C}}{{\rm{u}}^{2 + }}$
Now, we have to calculate the moles of zinc. The formula to calculate the number of moles is,
Number of moles$ = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar Mass}}}}$
The mass of zinc is given as 50 g and the molar mass of zinc is 65.4 g/mol.
Moles of zinc=$\dfrac{{50}}{{65.4}} = 0.76 < 1$
Therefore, zinc is consumed completely first.
Now, we have to calculate the charge required by 50 g zinc.
65.4 g of Zn (1 mole Zn) requires charge= $2 \times 96500\,{\rm{C}}$ (Two electrons involved)
Therefore, 50 g of zinc requires charge$ = \dfrac{{2 \times 96500}}{{65.4}} \times 50{\rm{C = 147554}}\,{\rm{C}}$
So, the charge (Q) of the reaction is 147554 C.
Now, we have to calculate time using the formula, $t = \dfrac{Q}{I}$. The value of Q is 147554 C and I is 1 ampere (C/sec)
$t = \dfrac{{147554\,{\rm{C}}}}{{1\,C/s}} = 147554\,s$
Now, we have to convert time to hour.
$ \Rightarrow 147554\,s = \dfrac{{147554}}{{3600}}{\rm{hour}}$
$ \Rightarrow 147554\,s = 40.98\,{\rm{hour}} \approx {\rm{41}}\,{\rm{hour}}$
Therefore, the cell runs upto 41 hours.

So, the correct answer is Option b.

Note: The galvanic cell consists of two half cells which are joined by a salt bridge that allows the passing of ions between the two sides to maintain electroneutrality. The conversion of chemical energy of a spontaneous reaction into electrical energy takes place in the galvanic cell.