Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31
Answer:
H2
Explanation:
In the first place it is necessary to consider that these two elements will be their diatomic form initially, that is, H2 and N2.
first we should check the equilibrium constant Kp tables in this case at a temperature of 3000K
Value for dissociation reaction of H2 in Kp = -3.685
Value for dissociation reaction of N2 in Kp = -22.359
Equilibrium constant for H2 dissociation is higher than N2 dissociation. so for this comparation H2 is more likely to dissociate.
The balanced chemical equation will tell you the molar ratios in which hydrogen and oxygen react to form water:
2H2 + O2 ----> 2H2O
Then the ratio is 2 mol of H2 react with 1 mol of O2 and form 2 mol of H2O:
2 : 1 : 2
Now convert the given data into moles:
16 grams of H2 = 16 g / 2 g/mol = 8 mol
16 grams of O2 = 16 g / 16 g/mol = 1 mol.
Then you have that the 1 mol of oxygen will react with 2 moles of hydrogen and they will form 2 mol of H2O.
And you can calculate the mass of 2 moles of H2O by multiplying by its molar weigth: 18 g/mol.
2 mol H2O * 18 g/mol = 36 grams of water.
Then, the answer is 36 grams of H2O.
Energy is absorbed to break bonds, and released when bonds are formed.
