solve 29 The charge required for the reduction of 1 mol of Cr2O72- to Cr3+ isa)96500 Cb) 1 93×105Cc) - Chemistry -

Question

solve 29
The   charge   required   for   the  
reduction   of   1   mol   of   Cr 2 O 7 2
-   to   Cr 3 +   is a ) 96500   C b )   1
93 × 10 5 C c )   5 79 × 10 5 C d )   2 895
× 10 5 C

Hina Parveen · Accepted Answer

Dear Student,
The change can be shown as

Cr2O72-→Cr3+
Cr changes from +6 to+3 hence there is a change of
3eâ€‹-
charge = n×F
so charge = 3F
we know 1F=96500
so charge needed = 289500 or 2 98×105C
hence correct option is (d)
Regards!

solve 29. The charge required for the reduction of 1 mol of Cr 2 O 7 2 - to Cr 3 + is a ) 96500 C b ) 1 . 93 × 10 5 C c ) 5 . 79 × 10 5 C d ) 2 . 895 × 10 5 C

solve 29.
$The charge required for the reduction of 1 mol of {Cr}_{2} {O_{7}}^{2 -} to {Cr}^{3 +} is a) 96500 C b) 1.93 \times 10^{5} C c) 5.79 \times 10^{5} C d) 2.895 \times 10^{5} C$