Answer:
0.48 V
Explanation:
Zn(s) ------------> Zn^2+(aq) + 2e. Oxidation half equation (-0.76V)
Co^2+(aq) + 2e-----------> Co(s). Reduction half equation (-0.28)
Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation
Zinc is the anode while cobalt is the cathode.
E°cell= E°cathode - E°anode
E°cell= -0.28-(-0.76)= 0.48 V
The accepted model of the atom was changed.
Answer:
Look at the image pls. The question is there
:
Answer:
2KBr + MgF₂ –> 2KF + MgBr₂
The coefficients are: 2, 1, 2, 1
Explanation:
KBr + MgF₂ –> KF + MgBr₂
The above equation can be balance as illustrated below:
KBr + MgF₂ –> KF + MgBr₂
There are 2 atoms of F on the left side and 1 atom on the right. It can be balance by writing 2 before KF as shown below:
KBr + MgF₂ –> 2KF + MgBr₂
There 2 atoms of K on the right side and 1 atom on the left side. It can be balance by writing 2 before KBr as shown below:
2KBr + MgF₂ –> 2KF + MgBr₂
Now, the equation is balanced.
The coefficients are: 2, 1, 2, 1
