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प्रश्न
Calculate \[\ce{E^0_{cell}}\] for the following cell.
\[\ce{Cr_{(s)} | Cr^{3+}_{( aq)} || Fe^{2+}){( aq)} | Fe_{(s)}}\]
Given: `"E"_("Cr"^(3+)//"Cr")^0` = −0.74 V,
`"E"_("Fe"^(2+)//"Fe")^0` = −0.44 V
पर्याय
−1.18 V
+1.18 V
+0.3 V
−0.3 V
MCQ
उत्तर
+0.3 V
Explanation:
Cr is anode and Fe is a cathode.
\[\ce{E^0_{cell} = E^0_{cathode} - E^0_{anode}}\]
= −0.44 − (−0.74)
= +0.3 V
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Electrode Potential and Cell Potential
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