Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Answer:
B
B
A
C
D
Explanation:
I think dont take my word for though
Answer:
The reaction will move to the left.
Explanation:
<em>Ba(OH)₂ = Ba²⁺ + 2OH⁻,</em>
<em>Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.</em>
- If H⁺ ions are added to the equilibrium:
H⁺ will combine with OH⁻ to form water.
<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>
<em />
<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.
- The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.
<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>
