When the following redox equation is balanced with smallest whole–number coefficients, the coefficient for ReO41- will be _____.
Au(s) + ReO4–(aq) --> Re(s) + Au3+(aq) (acidic solution)
A. |
2 |
|
B. |
3 |
|
C. |
7 |
|
D. |
8 |
|
E. |
None of these choices is correct. |
Au(s) + ReO4–(aq) --> Re(s) + Au3+(aq) (acidic solution)
first half
Au ..............>Au^+3
Au ..............>Au^+3 +3e- ...................1
second half
ReO4^- .............................> Re
ReO4^- .............................> Re + 4H2O
ReO4^- + 8H+ + 7e-.............................> Re + 4H2O ...............2
do eq.1*7 + eq.*3
3ReO4^- + 24H+ + 7Au............................> 3Re + 12 H2O + 7Au^+3
so option B is correct
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