Use the References to access important values if needed for this question. Enter
ID: 635682 • Letter: U
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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 remainingExplanation / 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
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