1828 United States presidential election in Ohio explained

See main article: 1828 United States presidential election.

Election Name:1828 United States presidential election in Ohio
Country:Ohio
Type:presidential
Ongoing:no
Previous Election:1824 United States presidential election in Ohio
Previous Year:1824
Next Election:1832 United States presidential election in Ohio
Next Year:1832
Election Date:October 31 – December 2, 1828
Image1:Andrew Jackson.jpg
Nominee1:Andrew Jackson
Party1:Democratic Party (United States)
Home State1:Tennessee
Running Mate1:John C. Calhoun
Electoral Vote1:16
Popular Vote1:67,595
Percentage1:51.60%
Nominee2:John Quincy Adams
Party2:National Republican Party (United States)
Home State2:Massachusetts
Running Mate2:Richard Rush
Electoral Vote2:0
Popular Vote2:63,394
Percentage2:48.40%
Map Size:200px
President
Before Election:John Quincy Adams
Before Party:Democratic-Republican
After Election:Andrew Jackson
After Party:Democratic Party (United States)

The 1828 United States presidential election in Ohio took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose 16 representatives, or electors to the Electoral College, who voted for President and Vice President.

Ohio voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won Ohio by a narrow margin of 3.2%. This was the first election of the Second Party System, and as such the first election in which Ohio voted for a candidate of a party other than the Democratic-Republicans.

Results

1828 United States presidential election in Ohio[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticAndrew Jackson67,59751.60%16
National RepublicanJohn Quincy Adams (incumbent)63,39648.40%0
Totals130,993100.0%16

See also

Notes and References

  1. Web site: 1828 Presidential General Election Results - Ohio. U.S. Election Atlas. 28 February 2013.