1820 United States presidential election in Ohio explained

See main article: 1820 United States presidential election.

Election Name:1820 United States presidential election in Ohio
Country:Ohio
Type:presidential
Ongoing:no
Previous Election:1816 United States presidential election in Ohio
Previous Year:1816
Next Election:1824 United States presidential election in Ohio
Next Year:1824
Election Date:November 1 – December 6, 1820
Nominee1:James Monroe
Party1:Democratic-Republican Party
Home State1:Virginia
Running Mate1:Daniel D. Tompkins
Electoral Vote1:8
Popular Vote1:7,164
Percentage1:76.38%
Nominee2:John Quincy Adams
Party2:Democratic-Republican Party
Home State2:Massachusetts
Electoral Vote2:0
Popular Vote2:2,215
Percentage2:23.62%
President
Before Election:James Monroe
Before Party:Democratic-Republican Party
After Election:James Monroe
After Party:Democratic-Republican Party

The 1820 United States presidential election in Ohio took place between November 1 and December 6, 1820, as part of the 1820 United States presidential election. Voters chose 8 representatives, or electors to the Electoral College, who voted for President and Vice President.

Ohio re-elected incumbent Democratic-Republican Party President James Monroe by a large margin. Although Monroe ran unopposed, John Quincy Adams received a minority of the vote as an opposition candidate. This is the most recent time a party won every county in the state.

Results

1820 United States presidential election in Ohio[1]
PartyCandidateVotesPercentageElectoral votes
Democratic-RepublicanJames Monroe (incumbent)7,16476.38%8
Democratic-RepublicanJohn Quincy Adams2,21523.62%0

See also

Notes and References

  1. Book: Ohio. Secretary of State. . 1909 . Ohio election statistics: the general election. . 3 . Springfield, Ohio . The Springfield Publishing Company, State Printers . 5470967 . 2019-04-20.