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1828 United States presidential election in Ohio

From Wikipedia, the free encyclopedia

1828 United States presidential election in Ohio

← 1824 October 31 – December 2, 1828 1832 →
Andrew Jackson.jpg
Nominee Andrew Jackson John Quincy Adams
Party Democratic National Republican
Home state Tennessee Massachusetts
Running mate John C. Calhoun Richard Rush
Electoral vote 16 0
Popular vote 67,595 63,394
Percentage 51.60% 48.40%

President before election

John Quincy Adams

Elected President

Andrew Jackson

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 would mark the first time a Democrat won a statewide federal election in Ohio, and the first time a Democrat won Ohio in a presidential election.

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  • How Louisiana has voted in Every Presidential Election



1828 United States presidential election in Ohio[1]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson 67,597 51.60% 16
National Republican John Quincy Adams 63,396 48.40% 0
Totals 130,993 100.0% 16

See also


  1. ^ "1828 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 28 February 2013.

This page was last edited on 23 February 2021, at 17:44
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