? I believe that the cathode is the magnesium electrode making the anode the pla

Question

?

I believe that the cathode is the magnesium electrode making the anode the platinum one.

I know that emf is just E, but I'm getting messed up with the calculation from the nernst equation (which I assume is what I need to use).

What is the spontaneous cell reaction?

And for the final part (D), I just use the fact that at equilibrium, E=0 so E°=(0.0257V)/n *lnK?

7.11 An electrochemical cell consists of a half-cell in which a piece of platinum wire is dipped into a solution that is 2.0 M in KBr and 0.050 M in Br2. The other halfcell consists of magnesium metal immersed in a 0.38 M Mg t solution. (a) Which electrode is the anode and which is the cathode? (b) What is the emf of the cell? (c) What is the spontaneous cell reaction? (d) What is the equilibrium constant of the cell reaction? Assume that the temperature is 25oC

Explanation / Answer

anode --> oxidation ( loss of electrons) occurs here

cathode --> reduction (gain of electrons) occurs here

so,

Br2 --> 0, will go to Br-, which is -1 it agains electrons

it gains electrons, it is being reduced, so it is the CATHODE

Solid plate of Mg goes from Mg0 to Mg+2, it loses 2 electrons

then, Mg plate is oxidized --> ANODE

Br2(aq) + 2 e 2 Br +1.0873

Mg2+ + 2 e Mg(s) 2.372

The reaction:

Mg(s) + Br2(aq) <--> 2Br-(aq) + Mg+2(aq)

E°cell = Ered - Eox = 1.0873--2.372 = 3.4593 V

b)

EMF of the cell

Ecell = E°cell -0.0592/n*log(Q)

Q = [Br-]^2 * [Mg+2] / [Br2]

Q = (2^2) * ( 0.38) / (0.05)= 30.4

Ecell = E°cell -0.0592/n*log(Q)

Ecell = 3.4593 -0.0592/2*log(30.4)

Ecell = 3.4154 V

c) this must be spontaneous, since it goes in the forward direction:

Mg(s) + Br2(aq) <--> 2Br-(aq) + Mg+2(aq)

d)

equilbirium constant is given for:

dG = -RT*lnK

dG = -nF*E°cell

RTln(K) = nF*Ecell

lnK = (nF)/(RT) * E°cell = (2*96500)/(8.314*298) * 3.4593 = 269.47

K = exp(269.47) = 1.069*10^117

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