Test Index

CBSE Class 12 Chemistry 2023 All Sets Solved Paper

© examsnet.com
Question : 3 of 18
Marks: +1, -0
A voltaic cell is made by connecting two half cells represented by half equations below:
Sn(aq)2++2e−→Sn(s)E∘=−0.14 V\mathrm{Sn}^{2+}_{(aq)}+2e^-\rightarrow\mathrm{Sn}_{(s)}E^\circ=-0.14\,\mathrm{V}
Fe(aq)3++e−→Fe(aq)2+E∘=+0.77 V\mathrm{Fe}^{3+}_{(aq)}+e^-\rightarrow\mathrm{Fe}^{2+}_{(aq)}E^\circ = +0.77\,\mathrm{V}
Which statement is correct about this voltaic cell?
Solution:  
Sn(aq)2++2e−→Sn(s)\mathrm{Sn}^{2+}_{(aq)}+2e^-\rightarrow\mathrm{Sn}_{(s)} E∘=−0.14 VE^\circ =-0.14\,\mathrm{V} ____(i)
2fe(aq)+3+2e−→2fe(aq)+22\mathrm{fe}^{+3}_{(aq)}+2e^-\rightarrow2\mathrm{fe}^{+2}_{(aq)} E1∘=+0.77 VE^\circ_1=+0.77\,\mathrm{V} ____(ii)
Sn(s)→Sn(aq)+2+2e−\mathrm{Sn}_{(s)}\rightarrow\mathrm{Sn}^{+2}_{(aq)}+2e^- E2∘=0.14E^\circ_2=0.14 _____(iii)
On adding equation (ii) and (iii)
2Fe(aq)+3+Sn(s)→2Fe(aq)+2+Sn+22\mathrm{Fe}^{+3}_{(aq)}+\mathrm{Sn}_{(s)}\rightarrow2\mathrm{Fe}^{+2}_{(aq)}+\mathrm{Sn}^{+2}, E∘E^\circ
−2FE∘=2FE1∘+2F-2FE^\circ=2FE^\circ_1+2F E2∘E^\circ_2
E∘=0.77+0.14E^\circ=0.77+0.14
E∘=0.91 VE^\circ=0.91\,\mathrm{V}
∵E∘>0\because E^\circ>0
∴ Sn is oxidised and the voltage at the cell is 0.91 V.
© examsnet.com
Go to Question:
Prev Question
Next Question