A cell is prepared by dipping a copper rod in 1M CuSO4 solution and a nickel rod in 1M NiSO4 solution. The standard reduction potentials of copper and nickel electrodes are +0.34 V and -0.25 V respectively.
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To determine the cell potential and the reaction that occurs in this electrochemical cell, we need to compare the standard reduction potentials of the two electrodes. The cell potential can be calculated using the Nernst Equation:
Ecell = E°cell - (0.0592/n) * log(Q)
where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred in the balanced equation
Q is the reaction quotient
Now, let's determine the cell potential and the reaction that occurs in this cell.
1. Write the balanced equation for the reaction at each electrode:
Copper electrode: Cu2+(aq) + 2e- -> Cu(s)
Nickel electrode: Ni2+(aq) + 2e- -> Ni(s)
2. Determine the value of n, the number of electrons transferred in the balanced equations. In both cases, it is 2.
3. Calculate the cell potential using the standard reduction potentials:
Ecell = E°cathode - E°anode
E°cathode = reduction potential of the cathode (nickel electrode) = -0.25 V
E°anode = reduction potential of the anode (copper electrode) = +0.34 V
Ecell = -0.25 V - (+0.34 V)
Ecell = -0.59 V
So, the cell potential is -0.59 V.
4. Determine the reaction that occurs in the cell based on the signs of the standard reduction potentials. Since the cell potential is negative, the reaction is not spontaneous as written. Therefore, the reverse reaction is favored:
Cu(s) + Ni2+(aq) -> Cu2+(aq) + Ni(s)
This means that copper metal oxidizes into copper ions (Cu2+) and nickel ions (Ni2+) reduce into nickel metal.
In summary, the cell potential of this electrochemical cell is -0.59 V, and the reaction that occurs is the oxidation of copper and the reduction of nickel.