The cell potential for Zn|Zn²⁺(aq)||Snˣ⁺| Sn is 0.801 V at 298 K. The reaction quotient for the above reaction is 10⁻². The number of electrons involved in the given electrochemical cell reaction is ____________.
( Given : E_(Zn²⁺ | Zn)ᵒ=-0.763 V, E_(Snˣ⁺ | Sn)ᵒ=+0.008 V and (2.303 RT)/F=0.06 V)

Solution

Chemistry ma’am · 4 steps

Here n is the only unknown in the Nernst equation, so put in everything else and solve for it.

1. Standard cell potential. Zn on the left is the anode and Sn on the right is the cathode. So E°_(cell) = E°_(cathode) - E°_(anode) = 0.008 - (-0.763) = 0.771 V.

   E_(cell) = E°_(cell) - 0.06/n log Q

2. Put in numbers. With Q = 10⁻², log Q = -2. This gives 0.801 = 0.771 - 0.06/n(-2) = 0.771 + 0.12/n.
3. Solve for n. 0.12/n = 0.801 - 0.771 = 0.03, so n = 0.12/0.03 = 4. This means tin is Sn⁴⁺, and the cell reaction is 2Zn + Sn⁴⁺ → 2Zn²⁺ + Sn.
4. So. The number of electrons involved in the cell reaction is 4.

The idea to keep. A Q less than 1 makes log Q negative, so E_(cell) comes out above E°_(cell). Use that check to catch a sign error before you solve for n.

Answer from the NTA final answer key. Solution by PrepWiser.