Calculate E for the following electrochemical cell at 25 degree C Pt(s)|Sn^2+ (aq, 0.5 M), Sn^4+...

Question

Question

science chemistry

Answer 1

Answer #1

Sn is anode and AgI is cathode.

So, oxidation occurs at anode and reduction at cathode.

E^o = E_anode + E_cathode

= (-0.15) + (-0.15) = -0.30 V

EMF = E^o - 0.06/n log [I-]²[Sn⁴⁺]/[Sn²⁺]

= -0.30 - 0.06/2 log (0.15)²(0.50)/(0.50)

= -0.30 - 0.03 log (0.15)²

= -0.30 - 0.03 log (0.0225)

= -0.30 - (0.03)(-1.65)

= -0.30 + 0.05 = -0.25 V

E of the cell is -0.25 V

Answer 2

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