Ecell 0=Ecathode −Eanode =1.33−(−0.44)=1.33+0.44=+1.77 V
The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is Fe(aq)2++2e−→Fe(s),E∘=−0.44 VCr2O72−(aq)+14H++6e−→2Cr3++7H2O,E0=+1.33 V
Held on 30 Apr 2023 · Verified 9 Jul 2026.
+0.01 V
+0.89 V
+1.77 V
+2.65 V
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