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An electrochemical cell is based on the following two half-reactions: oxidation: Sn(s)→Sn2+(aq, 2.00 M )+2e− reduction: ClO2(g, 0.110 atm )+e−→ClO−2(aq, 2.00 M )

User Hollden
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For the electrochemical cell with the given half-reactions and concentrations, the calculated cell potential using the Nernst equation is 0.98 V.

In this electrochemical cell, involving the overall reaction "Sn(s) + 2ClO2(g) → Sn2+(aq) + 2ClO2−(aq)," the standard cell potential (E^0_cell) is determined by the standard reduction potentials of the cathode (E^0_cathode) and the anode (E^0_anode).

Given :

E^0_cathode = 0.93 V and E^0_anode = -0.13 V, the standard cell potential is calculated as:

E^0_cell = E^0_cathode - E^0_anode = 0.93 V - (-0.13 V) = 1.06 V

Using the Nernst equation:

E_cell = E^0_cell - (0.0592/n) * log(Q),

Where n is the number of electrons (2 in this case) and Q is the reaction quotient, we find :

Q = [Sn2+][ClO2-]^2/[ClO2]^2 using the provided concentrations.

The calculated value of Q is 328.89.

Substituting values into the Nernst equation:

E_cell = 1.06 V - (0.0592/2) * log(328.89)

Solving this expression yields E_cell = 0.98 V.

The question probable may be:

An electrochemical cell is based on the following two half-reactions:

Ox: Sn(s)→Sn2+(aq, 1.85 M )+2e−

Red: ClO2(g, 0.150 atm )+e−→ClO−2(aq, 2.00 M )

Compute the cell potential at 25 ∘C.

Ecell =_______V

User DEEPTHI MUKUNDAN
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