A voltaic (galvanic) cell contains two half-cells. The anode is a manganese rod

Question

A voltaic (galvanic) cell contains two half-cells. The anode is a manganese rod placed in 0.500 L of 0.113 M MnS04. The cathode is a cadmium rod placed in 0.500 L of 0.0115 M CdS04. The starting cell potential is +0.753 V at 25°C. Determine the cell potential when half of the Cd2+ ions in this cell have been consumed. Did you recall the relationship between the cell potential, standard cell potential, and the reaction quotient? Did you recall how to determine the moles of electrons transferred during an oxidation-reduction reaction? Did you recall how to use stoichiometric ratios to calculate changes in molar quantities during a reaction

Explanation / Answer

Mn2+ + 2e-     ---->     Mn(s)    E0 = -1.185 v

Cd2+ + 2e-    ------>    Cd(s)    E0 = -0.403 v

FROM the data,

E0cell = E0c - E0a

       = -0.403 - (-1.185)

       = 0.782 v

Cell reaction : Mn(s) +Cd2+(aq) ----> Mn2+ (aq) + Cd(s)

   Ecell = E0cell - 0.0591/nlog[[Mn2+]/[Cd2+]]

0.753 = 0.782 -(0.0591/n)log(0.113/0.0115)

   n = no of e- transfered = 2

[Mn2+] = 0.111+0.00575 = 0.11675 M

[Cd2+] = 0.0115/2 = 0.00575 M

Ecell = 0.782-(0.0591/2)log(0.11675/0.00575)

         = 0.7433 v

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