Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions). Part A Label the anode and cathode, indicate the direction of electron flow.
Drag the appropriate labels to their respective targets.
Part B Write an equation for the half-reaction occurring at each electrode.
Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.
Part C What minimum voltage is necessary to drive the reaction?

Oxidation half equation:

Sn (s) ------> Sn²⁺ (aq) + 2e

Reduction half equation:

Mn²⁺ (aq) + 2e ----> Mn (s)

The minimum voltage necessary to drive the reaction is 1.055 V  

The setup of the electrolytic cell consists of a manganese and tin rods dipped in separate containers containing solutions of manganese chloride and tin chloride as electrolytes. A salt bridge is used to connect both half cells and in order to ensure ionic balance.

Since Mn²⁺ is reduced to Mn, the manganese rod serves as the cathode.

As Sn is oxidized to Sn²⁺, the tin rod serves as the anode.

The cathode is connected to the anode (negative terminal) of the battery, while the anode is connected to the cathode of the battery (positive terminal).

Electrons flows from  the anode to the cathode.

A battery is connected to the setup to drive the process from being non-spontaneous to spontaneity.

The setup is shown in image attached below.

Oxidation half equation:

Sn (s) ------> Sn²⁺ (aq) + 2e

Reduction half equation:

Mn²⁺ (aq) + 2e ----> Mn (s)

Cell voltage = E°cathode - E°anode

E°cathode= -1.195 V

E°anode= -0.140 V

Cell voltage= -1.195 V - (-0.140V)

Cell voltage= -1.055 V

Therefore, minimum voltage necessary to drive the reaction is 1.055 V

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-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.