A battery requires at least two electrodes, the anode at which oxidation occurs, and the cathode at which reduction occurs
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(Question and Answer): Can you please show me how to solve 7,8,9 using the values in 6. the question is , Assume that
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For Zn + Cu cell
Reduction half reaction
Oxidant + n e– – – – – > Reductant . . . Eo
Cu2+ + 2 e – – – – – > Cu . . . Eo = 0.339 V
For Zn + Fe cell
Fe2+ + 2 e – – – – – > Fe . . . Eo = – 0.44 V
For Zn + Ni cell
Ni2+ + 2 e – – – – – > Ni . . . Eo = – 0.25 V
For Zn + I2/I- cell
I2 + 2 e – – – – – > I2- . . . Eo = 0.54 V
Similarly you can write the reduction reactions for all cells.
Ans 8 and 9-
the most negative number, indicating that it is the strongest reducing agent as shown in the table below.
For ex- Lithium is the strongest reducing agent.
The strongest oxidizing agent, the largest positive number for standard electrode potential.
For ex- fluorine is the strongest oxidising agent.
Among the given cells
Strongest oxidizing agent is I2
Strongest reducing agent is Fe.