How many electrons are involved in the following redox reaction? `Cr_(2)O_(7)^(2-)` + `Fe^(2+)`+ `C_(2)O_(4)^(2-)` `to` `Cr^(3+)`+`Fe^(3+)`+`CO_(2)`
(Unbalanced)
(A)3
(B)4
(C)6
(D)5

How many electrons are involved in the following redox reaction? Cr_(2)O_(7)^(2-)+Fe^(2+)+C_(2)O_(4)^(2-)toCr^(3+)+Fe^(3+)+CO_(2)

In the following redox reactionn, Cr_(2)O_(7)^(2-) + Fe^(2+) to Fe^(3+) + Cr^(3+) 1 mole of Cr_(2)O_(7)^(2-) oxidises.

The number of electrons donated from substance(s) getting oxidized to the substance(s) getting reduced in the chemical equaton for the following reaction is: Cr_2O_(7)^(2-) + Fe^(2+) + C_2O_(4)^(2-) rarr Cr^(3+) +Fe^(3+) + CO_2 (Unbalanced)

How many moles of electrons are involved in the conversion of 1 mol Cr_(2)O_(7)^(2-) into Cr^(3+) ion? Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O

How many mole of electrons are involved balancing the following equations: (a) H_(2)S+NO_(3)^(-)rarrS+NO (b) Mn(OH)_(2)+H_(2)O_(2)rarrMnO_(2)+2H_(2)O (c ) Cr_(2)O_(7)^(2-)+Fe^(2+)+C_(2)O_(4)^(2-)rarrCr^(3+)+Fe^(3+)+CO_(2) (acid medium) (d) Br_(2)+OH^(-)rarrBrO_(3)^(-)+Br^(-) (e ) The compound P_(4)S_(3) is oxidised by nitrate ions acid medium to give phosphoric acid, sulphate ions and nitric oxide (NO) . Write the balanced half reactions and the overall reaction.

Balance the following equation stepwise: Cr_(2)O_(7)^(2-) + Fe^(2+)++H^(o+)rarrCr^(3+) + Fe^(3+) + H_(2)O

Balance the following reaction by ion electron method. Cr_(2)O_(7)^(-2)+SO_(3)^(-2) rightarrow Cr^(3+)+SO_(4)^(-2)

In the following unbalanced redox reaction, Cu_(3)P+Cr_(2)O_(7)^(2-) rarr Cu^(2+)+H_(3)PO_(4)+Cr^(3+) Equivalent weight of H_(3)PO_(4) is