Asked by RZeal
                Calculate the Ecell for this reaction
3HClO2(aq) +2Cr3+(aq) + 4H2O(l) -> 3HClO(aq) + (Cr2O7)2–(aq) + 8H+(aq)
I thought Cr + e -> Cr2O7 is the cathode
and 3HClO2 -> 3HClO + e- is the anode.
Using Ecell = Cathode - Anode
I get -0.31, but the answer I checked is supposed to be positive.
If someone can explain me the steps again that'll be great, thanks
            
        3HClO2(aq) +2Cr3+(aq) + 4H2O(l) -> 3HClO(aq) + (Cr2O7)2–(aq) + 8H+(aq)
I thought Cr + e -> Cr2O7 is the cathode
and 3HClO2 -> 3HClO + e- is the anode.
Using Ecell = Cathode - Anode
I get -0.31, but the answer I checked is supposed to be positive.
If someone can explain me the steps again that'll be great, thanks
Answers
                    Answered by
            DrBob222
            
    See your other post above.
    
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