In the reaction:
`Cr_(2)O_(7)^(2-) + 14H^(o+) + 6I^(Ө) rarr2Cr^(3+) + 3H_(2)O + 3I_(2)`
Which element is reduced?

In the reaction Cr_(2) O_(7)^(2-) + 14 H^(+) + 6I^(-) to 2 Cr^(3+) + 3H_(2)O + I_(2) Wchich element is reduced?

In the following reaction Cr_(2) O_(7)^(2-) + 14H^(+) + 6I^(-) rarr 2Cr^(3+) + 7H_(2)O + 3I_(2) Which element gets reduced ?

In the reaction , Cr_2O_7^(2-) + 14H^+ +6I → 2Cr^(3+) + 7H_2 O +3I_(2 which element is reduced ?

Consider the reaction, Cr_(2) O_(7) ^(2-) + 14 H^(+) + 6e^(-) rarr 2Cr^(3+) + 7 H_(2) O What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr_(2)O_(7)^(2-) ?

Balance the equation : Cr_3O_7^(2-) + H^+ + I^- rarr Cr^3 + I2 +H_2O

Which substance is serving as a reducing agent in the following reaction ? Cr_(2) O_(7)^(2-) + 14H^(+) + 3Ni rarr 2Cr^(3+) + 7H_(2)O + 3Ni^(2+)