Question

Calculate the cell potential of a copper electrode immersed in (i) 0.0750 M in NaCI and saturated CuCI 0.0400 M in NaOH and saturated with Cu(OH)2

          Calculate the cell potential of a copper electrode immersed in (i)
0.0750 M in NaCI and saturated CuCI
0.0400 M in NaOH and saturated with Cu(OH)2
        

Added by Kristy A.

Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Calculate the cell potential of a copper electrode immersed in (i) 0.0750 M in NaCI and saturated CuCI 0.0400 M in NaOH and saturated with Cu(OH)2
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An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10$M \mathrm{NaOH}$ that is saturated with $\mathrm{Cu}(\mathrm{OH})_{2},$ what is the cell potential at $25^{\circ} \mathrm{C} ?\left[\text { For } \mathrm{Cu}(\mathrm{OH})_{2}\right.$ $K_{\mathrm{sp}}=1.6 \times 10^{-19} . ]$

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Transcript

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00:01 So from the question here, we are to calculate the potential of the copper electrode immersed in a 0 .0 440m, c u .03 to nb, 0 .075m in nacl and saturated with cul.
00:20 So let's look at the first one.
00:22 So before that we have the next equation which is given here and this is going to help us getting the answers the next equation so here the e not is a standard electric potential the r is a gas constant the t is a temperature in kelvin the is the number of most of electron the f is a fire day constant and the q is the reaction quotient the q is a reaction quotient so at 25 degrees celsius the equation becomes e equals e equals or e degree minus 0 .0592 over n log q so let's look at the first one the a so with the a in focus we have with the a in focus we have two electrons plus c u two plus goes to c u which is solid where the e knot is equal to plus 0 .337v.
01:48 So using the next equation that i mentioned earlier, using the next equation, remember it's here, using the next equation we have e equals e0 minus 0 .0592 all over and log q.
02:26 And so this becomes our n is giving us 2.
02:41 So we have e equals, sorry, e to the power 0 plus 0 .337v.
02:53 So in all, we have something like this...
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