Question

A Galvanic cell is constructed as follows. A half- cell consists of a platinum wire immersed in a solution containing 1.0 M of $Sn^{2 +}$and 1.0M of $Sn^{4 +}$and another half- cell has a thallium rod immersed in a 1.0 M solution of $Tl^{+}$

Given $Sn^{4 +}(aq) + 2e \rightarrow Sn^{2 +}(aq);E^{0} = + 0.13V$

and$Tl^{+}(aq) + e \rightarrow Tl(s);E^{0} = - 0.34V,$

What is the cell voltage if the $Tl^{+}$concentration is increased tenfold

• A. 0.411V
• B. 4.101 V
• C. 0.492 V
• D. 0.222 V

Answer 1

Answer: (A)

0.411V

Answer 2

The cell is represented as

$Tl(\overset{- 0.34V}{s)|Tl^{+}}(1.0M)||Sn^{4 +}(1.0M),Sn^{+ 2}(1.0\overset{+ 0.13V}{M)|Pt}$

The cell reaction is [$Tl(s) \rightarrow Tl^{+} + e \times 2$]

$Sn^{4 +} + 2e^{-} \rightarrow Sn^{2 +}$

Overall reaction : $2Tl(s) + Sn^{4 +} \rightarrow 2Tl^{+} + Sn^{2 +}$ [Concentration $(Tl^{+}) = (10M)$, $(Sn^{2 +}) = 1M$, $(Sn^{4 +}) = 1M$]

$E = (E_{R}^{0} - E_{L}^{0}) - \frac{0.0592}{2}\log\frac{\lbrack Tl^{+}\rbrack^{2}\lbrack Sn^{2 +}\rbrack}{\lbrack Sn^{4 +}\rbrack}$

$= 0.47V - 0.0296\log(10)^{2}\lbrack Tl^{+}$concentration increases tenfold]

= 0.47 – 0.0592 = 0.411V