Question

A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:

Answer 1

Answer: The cell voltage of the given chemical reaction is 2.74 V

Explanation:

For the given chemical reaction:

$Sn^{2+}(aq.)+Ba(s)\rightarrow Ba^{2+}(aq.)+Sn(s)$

Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.

Oxidation half reaction:  $Ba(s)\rightarrow Ba^{2+}(aq.)+2e^-;E^o_{Ba^{2+}/Ba}=-2.90V$

Reduction half reaction:  $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s);E^o_{Sn^{2+}/Sn}=-0.14V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.14-(-2.90)=2.76V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]}{[Ba^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = +2.76 V

n = number of electrons exchanged = 2

$[Ba^{2+}]=5.15M$

$[Sn^{2+}]=1.59M$

Putting values in above equation, we get:

$E_{cell}=2.76-\frac{0.059}{2}\times \log(\frac{1.59}{5.15})\\\\E_{cell}=2.74V$

Hence, the cell voltage of the given chemical reaction is 2.74 V