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Question & Answer: A concentration cell based on the following half reaction at 305 K Mg2+ + 2 e- → Mg SRP =…..

A concentration cell based on the following half reaction at 305 K Mg2+ + 2 e- → Mg SRP = -2.37 V has initial concentrations of 1.27 M Mg2+, 0.255 M Mg2+, and a potential of 0.02110 V at these conditions. After 10 hours, the new potential of the cell is found to be 0.009984 V. What is the concentration of Mg2+ at the cathode at this new potential?

Expert Answer

Answer

$E_{cell} = dfrac {0.0591}{n}log dfrac {[Mg^{2+}] _{cathode}}{[Mg^{2+}] _{anode}}$

$0.02110 V = dfrac {0.0591}{2} log dfrac { 1.27 }{0.255}$

$0.02110 : V = 0.02110 : V$

Thus electrode with $[Mg^{2+}] = 1.27 : M$ is cathode and the electrode with $[Mg^{2+}] = 0.255 : M$ is anode.

At cathode $[Mg^{2+}]$ will decrease to $left ( 1.27-x right ) : M$ . At anode $[Mg^{2+}]$ will increase to $left ( 0.255 + x right ) : M$ .

After 10 hours, the new potential of the cell is found to be 0.009984 V.

$E_{cell} = dfrac {0.0591}{n}log dfrac {[Mg^{2+}] _{cathode}}{[Mg^{2+}] _{anode}}$

$0.009984 V = dfrac {0.0591}{2} log dfrac { 1.27-x }{0.255+x}$

$log dfrac { 1.27-x }{0.255+x} = 0.3379$

$dfrac { 1.27-x }{0.255+x} =2.177$

$1.27-x =2.177 left ( 0.255+x right )$

$1.27-x = 0.555+2.177x$

$1.27 = 0.555+3.177x$

$3.177x=0.7149$

$x=0.225$

$[Mg^{2+}] = 1.27-x = 1.27 - 0.225 = 1.05 : M$

This is the concentration of $Mg^{2+}$ at the cathode at this new potential

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