1. If half cell A has a standard reduction potential of E= -1.10 V and half cell B has a standard reduction potential of E= 0.65 V, which half-cell is the anode?
2. Which half cell has the higher potential energy?
3. Calculate Ecell for this voltaic cell.
A + e ==> A^- Eo = -1.10vb
B + e ==> B^- Eo = 0.65v
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For a spontaneous cell A must be reversed; therefore,
A^- ==> A + e E = 1.10
B + e ==> B^- E = 0.65
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A^- + B ==> A + B^-
Ecell = 1.10 + 0.65 = ?
The anode is where oxidation occurs and oxidation is the loss of electrons; therefore, A^- must be the anode.
To determine which half-cell is the anode, you need to compare the standard reduction potentials (E) of the half-cells.
1. The anode is the half-cell with the lower standard reduction potential (E). In this case, since the half-cell A has a standard reduction potential of E = -1.10 V and half-cell B has a standard reduction potential of E = 0.65 V, half-cell A would be the anode as it has the lower reduction potential.
To determine which half-cell has the higher potential energy:
2. The half-cell with a higher standard reduction potential (E) has a higher potential energy. So, in this case, the half-cell B with E = 0.65 V has higher potential energy compared to half-cell A with E = -1.10 V.
To calculate the cell potential (Ecell) for this voltaic cell:
3. Ecell is calculated by subtracting the reduction potential of the anode (Eanode) from the reduction potential of the cathode (Ecathode).
Ecell = Ecathode - Eanode
In this case, the reduction potential of half-cell B (cathode) is 0.65 V (given), and the reduction potential of half-cell A (anode) is -1.10 V (given).
Ecell = 0.65 V - (-1.10 V)
= 0.65 V + 1.10 V
= 1.75 V
So, the calculated Ecell for this voltaic cell is 1.75 V.