The EMF of the cell : Ni(s) | Ni2+ (1.0M) || Au 3+ (1.0M) | Au (s) is[ given EoNi2+/Ni -0.25V; E° Au3+ / Au = +1.50V ]
Question
The EMF of the cell : Ni(s) | Ni2+ (1.0M) || Au 3+ (1.0M) | Au (s) is[ given EoNi2+/Ni -0.25V; E° Au3+ / Au = +1.50V ]
Solution
To find the EMF of the cell, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
First, let's determine the half-reactions occurring at each electrode:
At the anode (left side): Ni(s) -> Ni2+(aq) + 2e-
At the cathode (right side): Au3+(aq) + 3e- -> Au(s)
Now, we can calculate the cell potential using the given standard reduction potentials:
E°cell = E°cathode - E°anode E°cell = E°Au3+/Au - E°Ni2+/Ni E°cell = +1.50V - (-0.25V) E°cell = +1.75V
Next, we need to calculate the reaction quotient, Q, using the concentrations of the species involved:
Q = [Ni2+]/[Au3+]
Since both concentrations are given as 1.0M, Q = 1.0/1.0 = 1.0
Now, we can substitute the values into the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Since the number of electrons transferred in the balanced equation is 2, n = 2:
Ecell = 1.75V - (0.0592/2) * log(1.0) Ecell = 1.75V - (0.0296) * log(1.0) Ecell = 1.75V - 0 Ecell = 1.75V
Therefore, the EMF of the cell is 1.75V.
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