Question

Balance the the following Redox Equation :

Answer 1

\[Cr _{2}O _{7}^{-2}+I ^{-} ------> Cr ^{+3} + I _{2}\]

Answer 2

i know all the steps but still cant understand

Answer 3

Spliting into 2 Partial Equations

Answer 4

first step ok

Answer 5

Reduction:-
\[Cr _{2} ^{+12} ---> 2Cr ^{+3}\]
Oxidation Number for Cr = 12

Answer 6

the total might be +12, but there's 2 ions to share the charge, so it's +6 each. You also don't want to split the ion up, keep it as Cr2O7

Answer 7

Reduction:-
\[2I^- --> I_2\]

Answer 8

sorry it was cr207 ---> 2Cr^+3

Answer 9

\[Cr_2O_7^{-2} \rightarrow 2Cr^{+3}\]and \[2I^{-1} \rightarrow I_2\]

Answer 10

Now balancing by adding H+ and H2O

Answer 11

balance O's first by adding H2O

Answer 12

then balance H+

Answer 13

this is the part i need help and there is further more part of electrons balancing

Answer 14

yes plz explain to me help plzz ...how to balance this by H+ and H2O

Answer 15

balance O's by adding water

then balance the waters by adding H+

then balance charge by adding electrons

Answer 16

Oh man!!! .....it was so easy ....

Answer 17

\[Cr_2O_7^{-2} \rightarrow 2Cr^{+3}\] balance Os with water\[Cr_2O_7^{-2} \rightarrow 2Cr^{+3} + 7H_2O\] balance H's with H+\[14H^{+1} + Cr_2O_7^{-2} \rightarrow 2Cr^{+3} + 7H_2O\] balance charge with electrons\[6e^{-1} + 14H^{+1} + Cr_2O_7^{-2} \rightarrow 2Cr^{+3} + 7H_2O\]

Answer 18

YES this is the part !!! i really suck at this part

Answer 19

how did u added 6e ?

Answer 20

in the next-to-bottom equation, the total charge of the reactants is +12. the total charge of all the products is +6. The only way to make those charges EQUAL, is to add 6 units of negative charge to the reactants

Answer 21

and what about the Reduction equation ?

Answer 22

the reduction is easier because there's no oxygens or hydrogens to balance. so look at the charges and put electrons where they would balance the charges

Answer 23

how did u get the total of reactants and total of products ? can u tell me ?

Answer 24

yeah its easy !! 2e are going to add on right side

Answer 25

2e- goes on the right side, correct

now multiply each half-reaction so that the electrons totally cancel when you add them back together

Answer 26

In oxidation equation how did u get the total charge of reactants and total charge of products ?

Answer 27

\[6e^{-1} + 14H^{+1} + Cr_2O_7^{-2} \rightarrow 2Cr^{+3} + 7H_2O\]and \[2I^{-1} \rightarrow I_2 + 2e^{-1}\] don't have the same number of electrons, so multiply the second half-reaction by 3, to make 6 electrons\[6I^{-1} \rightarrow 3I_2 + 6e^{-1}\]now add both half-reactions

Answer 28

yeah the next step of making electrons equal and then adding them is a piece of cake!!!

Answer 29

add up the charges of all the reactants, that's the total charge

Answer 30

what are the charges ? plz i need help only in this ELECTRON adding part

Answer 31

the H+ ions each have a charge of +1, and there are 14 of them

the Cr2O7 has a charge of -2, and there's only 1 of them

add +14 and -2 together, and you get +12

Answer 32

And what is the total charge of this \[2Cr^{+3} + 7H_2O\]

Answer 33

@JFraser ??

Answer 34

each Cr ion has a charge of +3, and there are 2 of them...

Answer 35

i am solving a basic equation right now :
\[Cr(OH)_3 --> CrO^-4\]
How to Balance by adding OH- and H2O ??

Answer 36

u said first balance O's by adding H2O ........

Answer 37

but this is difficult

Answer 38

hello u there ??

Answer 39

i'm checking your formulas, hang on

Answer 40

what happened ??

Answer 41

balancing in basic solutions isn't that much different from acidic solutions, but there are no H+ ions left over once the reactions are balanced