Answer: 1.19 V
Explanation:
Standard potential for an electrochemical cell is given by:
[tex]E^0{cell}[/tex] = standard electrode potential =[tex]E^0{cathode}-E^0{anode}[/tex]
The [tex]E^0[/tex] values have to be reduction potentials.
Given:
[tex]E^o_{Cd^{2+}/Cd}=-0.40V[/tex]
[tex]E^o_{Hg_2^{2+}/Hg}=0.79V[/tex]
[tex]Cd+Hg_2^{2+}\rightarrow Cd^{2+}+2Hg[/tex]
Cadmium is losing electrons , undergo oxidation and thus act as anode. mercury is gaining electrons, undergo reduction and thus acts as cathode.
[tex]E^0{cell}[/tex] = standard electrode potential =[tex]E^0{cathode}-E^0{anode}=0.79-(-0.40)=1.19V[/tex]
Thus the standard potential for an electrochemical cell with the cell reaction is 1.19 V.
The overall cell potential for this redox reaction : 1.19V
Cd + Hg₂²⁺ → Cd²⁺ + 2Hg
Cell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.
Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)
[tex]\large{\boxed {\bold{E ^osel = E ^ocatode -E ^oanode}}}[/tex]
or:
E ° cell = E ° reduction-E ° oxidation
(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)
The value of E⁰ cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode
In reaction:
Cd + Hg₂²⁺ → Cd²⁺ + 2Hg
Given:
Cd²⁺+ 2e⁻ → Cd (s) = –0.40V and
Hg²⁺ + 2e⁻ → 2Hg (l) = 0.79V
From the value of E ° cells, it can be seen that the higher E ° cells will act as positive poles or cathodes namely Hg₂²⁺
So the reaction happens
at the anode (oxidation reaction) Cd ------> Cd²⁺ + 2e⁻ E ° = +0.40 V
at the cathode (reduction reaction) Hg₂²⁺ + 2e− → 2Hg (l) = 0.79 V
E ° cell = E ° cathode -E ° anode = 0.79 V - (-0.40V) = 1.19V
The standard cell potential
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Keywords: cell potential, cathode, anode, reduction, oxidation
