Question & Answer: Can you please show me how to solve 7,8,9 using the values in 6. the question is , Assume that…..

Can you please show me how to solve 7,8,9 using the values in 6. the question is , Assume that the reduction potential for your most active metal is 0.0V. using your observed voltages, determine the reduction potentials for the other oxidized species investigated.B. Measurement of E Values 6. Cathode Voltage Cell Zn + Cu Zn + Fe Zn +Ni Anode 51 Cu Fe o 15 Cu + Ni Fe + Ni Ni12/1 Reduction Half-equation (oxidized form) + ne--> (reduced form) 7. Relative Reduction Potential Strongest oxidizing agent Strongest reducing agent 8. 9.

B. Measurement of E Values 6. Cathode Voltage Cell Zn + Cu Zn + Fe Zn +Ni Anode 51 Cu Fe o 15 Cu + Ni Fe + Ni Ni12/1 Reduction Half-equation (oxidized form) + ne–> (reduced form) 7. Relative Reduction Potential Strongest oxidizing agent Strongest reducing agent 8. 9.

Expert Answer

Answer

A battery requires at least two electrodes, the anode at which oxidation occurs, and the cathode at which reduction occurs

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.

Question & Answer: Can you please show me how to solve 7,8,9 using the values in 6. the question is , Assume that..... 1

Among the given cells

Strongest oxidizing agent is I2

Strongest reducing agent is Fe.

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