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Use the References to access important values if needed for this question. Enter

ID: 635682 • Letter: U

Question

Use the References to access important values if needed for this question. Enter electrons as e A voltaic cell is constructed in which the anode is a Mg Mg half cell and the cathode is a FelFe half cell. The half-cell compartments are connected by a salt bridge. (Use the lowest possible coefficients. Use the pul-down boxes to specify states such as (aq) or (3) If a box is not needed, leave it blank.) The anode reaction is: The cathode reaction is The net cell reaction is In the external circuit, electrons migrate? i? the MgMg2. electrode? the FeFe2. electrode. In the salt bridge, anions migrate t the FelFe compartment the MgMg compartment Submit Answer Retry Entire Group 9 more group attempts remaining

Explanation / Answer

At the anode - Oxidation

Ni(s) ---> Ni2+ (aq) + 2e

At Cathode - Reduction

F2 (g) + 2e -----> 2F-(aq)

net cell reaction

Ni(s) + F2(g) -----> 2F-(aq) + Ni2+(aq)

In the external circuit electrons migrates from Mg/Mg2+ electrode to Fe/Fe2+ electrode.

In the salt bridge anions migrate from Fe/Fe2+ compartment to the Mg/Mg2+ compartment.

At the anode - Oxidation

Mg(s) ---> Mg2+ (aq) + 2e

At Cathode - Reduction

net cell reaction

Fe2+ (aq) + 2e -----> Fe(s)

In the external circuit electrons migrates from Ni/Ni2+ electrode to F-/F2 electrode.

In the salt bridge anions migrate from  F-/F2 compartment to the Ni/Ni2+ compartment.

At the anode - Oxidation

Mn(s) ---> Mn2+ (aq) + 2e

At Cathode - Reduction

Br2 (g) + 2e -----> 2Br-(aq)

The net reaction

Mn(s) + Br2 (g) ----> 2Br-(aq) + Mn2+ (aq)

The cell voltage = E(redcution) - E(oxidation)

E = 1.080 - (-1.180) = 2.260V

At the anode - Oxidation

3Mg(s) ---> 3Mg2+ (aq) + 6e

At Cathode - Reduction

2Al3+(aq) + 6e -----> 2Al(s)

The net reaction

3Mg(s) + 2Al3+ (aq) ----> 2Al(s) + 3Mn2+ (aq)

The cell voltage = E(redcution) - E(oxidation)

E = -1.660 - (-2.370) = 0.71 V