Question

Calculate Ecell for the following electrochemical cell at 25 ºCPt (s) | H2 (g, 1.00 atm) | H+ (aq, 1.00 M) || Sn2+ (aq, 0.350 M) | Sn4+ (aq, 0.020 M) | Pt (s)The standard reduction potentials are as follows:Sn4+ (aq) + 2 e–à Sn2+ (aq) Eº = +0.13 V2 H+ (aq) + 2 e–à H2 (g) Eº = 0.00 V

Answer 1

Answer: The electrode potential of the cell is 0.093 V

Explanation:

The given chemical cell follows:

$Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)$

Oxidation half reaction: $H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V$

Reduction half reaction: $Sn^{4+}(aq,0.020M)+2e^-\rightarrow Sn^{2+(aq.,0.350M);E^o_{Sn^{4+}/Sn^{2+}}=0.13V$

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=0.13-(0.0)=0.13V$

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

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2[Sn^{2+}]}{[Sn^{4+}]}$

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

$[H^{+}]=1.00M$

$[Sn^{2+}]=0.350M$

$[Sn^{4+}]=0.020M$

Putting values in above equation, we get:

$E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})$

$E_{cell}=0.093V$

Hence, the electrode potential of the cell is 0.093 V